Question

A common way of removing Cr³⁺ from solution is precipitation of Cr(OH)₃(s). An environmental engineer tells you that adjusting the pH to 8.5 will decrease the total soluble Cr(III) to below 1 mg/L. Is this true? In other words, is the solubility of Cr(III) in the presence of Cr(OH)₃(s) at a pH of 8.5 less than 1 mg/L? the pK_spof Cr(OH)₃(s) is 30.22.

Note: This question is in 5th edition of Chemistry for Environmental Engineering and Science. Chapter 4, Question 86.

Answer 1

PH   = 8.5

POH = 14-PH

          = 14-8.5 = 5.5

-log[OH-] = 5.5

   [OH-]   = 10^-5.5   = 3.16*10^-6 M

P^Ksp     = 30.22

-logKsp = 30.22

Ksp   = 10^-30.22    = 6.025*10^-31

Cr(OH)3 -------> Cr⁺³ + 3OH^-

Ksp   = [Cr³⁺][OH-]³

6.025*10^-31 = [Cr³⁺]*(3.16*10^-6 )³

[Cr⁺³]    = 6.025*10^-31/(3.16*10^-6)³

             = 1.9*10^-4 mole/L

          = 1.9*10^-4 * 52g/L

          = 98.8*10^-4g/L

        = 9.88mg/L